I want expanding mean of col2 based on groupby('col1'), but I want the mean to not include the row itself (just the rows above it)
dummy = pd.DataFrame({"col1": ["a",'a','a','b','b','b','c','c'],"col2":['1','2','3','4','5','6','7','8'] }, index=list(range(8)))
print(dummy)
dummy['one_liner'] = dummy.groupby('col1').col2.shift().expanding().mean().reset_index(level=0, drop=True)
dummy['two_liner'] = dummy.groupby('col1').col2.shift()
dummy['two_liner'] = dummy.groupby('col1').two_liner.expanding().mean().reset_index(level=0, drop=True)
print(dummy)
---------------------------
here is result of first print statement:
col1 col2
0 a 1
1 a 2
2 a 3
3 b 4
4 b 5
5 b 6
6 c 7
7 c 8
here is result of the second print:
col1 col2 one_liner two_liner
0 a 1 NaN NaN
1 a 2 1.000000 1.0
2 a 3 1.500000 1.5
3 b 4 1.500000 NaN
4 b 5 2.333333 4.0
5 b 6 3.000000 4.5
6 c 7 3.000000 NaN
7 c 8 3.800000 7.0
I would have thought their results would be identical.
two_liner is the expected result. one_liner mixes numbers in between groups.
It took a long time to figure out this solution, can anyone explain the logic? Why does one_liner not give expected results?
You are looking for expanding().mean() and shift() within the groupby():
groups = df.groupby('col1')
df['one_liner'] = groups.col2.apply(lambda x: x.expanding().mean().shift())
df['two_liner'] = groups.one_liner.apply(lambda x: x.expanding().mean().shift())
Output:
col1 col2 one_liner two_liner
0 a 1 NaN NaN
1 a 2 1.0 NaN
2 a 3 1.5 1.0
3 b 4 NaN NaN
4 b 5 4.0 NaN
5 b 6 4.5 4.0
6 c 7 NaN NaN
7 c 8 7.0 NaN
Explanation:
(dummy.groupby('col1').col2.shift() # this shifts col2 within the groups
.expanding().mean() # this ignores the grouping and expanding on the whole series
.reset_index(level=0, drop=True) # this is not really important
)
So that the above chained command is equivalent to
s1 = dummy.groupby('col1').col2.shift()
s2 = s1.expanding.mean()
s3 = s2.reset_index(level=0, drop=True)
As you can see, only s1 considers the grouping by col1.
Related
I know codes forfilling seperately by taking each column as below
data['Native Country'].fillna(data['Native Country'].mode(), inplace=True)
But i am working on a dataset with 50 rows and there are 20 categorical values which need to be imputed.
Is there a single line code for imputing the entire data set??
Use DataFrame.fillna with DataFrame.mode and select first row because if same maximum occurancies is returned all values:
data = pd.DataFrame({
'A':list('abcdef'),
'col1':[4,5,4,5,5,4],
'col2':[np.nan,8,3,3,2,3],
'col3':[3,3,5,5,np.nan,np.nan],
'E':[5,3,6,9,2,4],
'F':list('aaabbb')
})
cols = ['col1','col2','col3']
print (data[cols].mode())
col1 col2 col3
0 4 3.0 3.0
1 5 NaN 5.0
data[cols] = data[cols].fillna(data[cols].mode().iloc[0])
print (data)
A col1 col2 col3 E F
0 a 4 3.0 3.0 5 a
1 b 5 8.0 3.0 3 a
2 c 4 3.0 5.0 6 a
3 d 5 3.0 5.0 9 b
4 e 5 2.0 3.0 2 b
5 f 4 3.0 3.0 4 b
I have the following Dataset:
col value
0 A 1
1 A NaN
2 B NaN
3 B NaN
4 B NaN
5 B 1
6 C 3
7 C NaN
8 C NaN
9 D 5
10 E 6
There is only one value set per group, the rest in Nan. What I want to do know, is fill the NaN with he value of the group. If a group has no NaNs, I just want to ignore it.
Outcome should look like this:
col value
0 A 1
1 A 1
2 B 1
3 B 1
4 B 1
5 B 1
6 C 3
7 C 3
8 C 3
9 D 5
10 E 6
What I've tried so far is the following:
df["value"] = df.groupby(col).transform(lambda x: x.fillna(x.mean()))
However, this method is not only super slow, but doesn't give me the wished result.
Anybody an idea?
It depends of data - if there is always one non missing value you can sorting and then replace by GroupBy.ffill, it working well if some groups has NANs only:
df = df.sort_values(['col','value'])
df["value"] = df.groupby('col')["value"].ffill()
#if always only one non missing value per group, fail if all NaNs of some group
#df["value"] = df["value"].ffill()
print (df)
col value
0 A 1.0
1 A 1.0
5 B 1.0
2 B 1.0
3 B 1.0
4 B 1.0
6 C 3.0
7 C 3.0
8 C 3.0
9 D 5.0
10 E 6.0
Or if there is multiple values and need replace by mean, for improve performace change your solution with GroupBy.transform only mean passed to Series.fillna:
df["value"] = df["value"].fillna(df.groupby('col')["value"].transform('mean'))
print (df)
col value
0 A 1.0
1 A 1.0
5 B 1.0
2 B 1.0
3 B 1.0
4 B 1.0
6 C 3.0
7 C 3.0
8 C 3.0
9 D 5.0
10 E 6.0
You can use ffill which is the same as fillna() with method=ffill (see docs)
df["value"] = df["value"].ffill()
Trying to groupby in pandas, then sort values and have a result column show what you need to add to get to the next row in the group, and if your are the end of the group. To replace the value with the number 3. Anyone have an idea how to do it?
import pandas as pd
df = pd.DataFrame({'label': 'a a b c b c'.split(), 'Val': [2,6,6, 4,16, 8]})
df
label Val
0 a 2
1 a 6
2 b 6
3 c 4
4 b 16
5 c 8
Id like the results as shown below, that you have to add 4 to 2 to get 6. So the groups are sorted. But if there is no next value in the group and NaN is added. To replace it with the value 3. I have shown below what the results should look like:
label Val Results
0 a 2 4.0
1 a 6 3.0
2 b 6 10.0
3 c 4 4.0
4 b 16 3.0
5 c 8 3.0
I tried this, and was thinking of shifting values up but the problem is that the labels aren't sorted.
df['Results'] = df.groupby('label').apply(lambda x: x - x.shift())`
df
label Val Results
0 a 2 NaN
1 a 6 4.0
2 b 6 NaN
3 c 4 NaN
4 b 16 10.0
5 c 8 4.0
Hope someone can help:D!
Use groupby, diff and abs:
df['Results'] = abs(df.groupby('label')['Val'].diff(-1)).fillna(3)
label Val Results
0 a 2 4.0
1 a 6 3.0
2 b 6 10.0
3 c 4 4.0
4 b 16 3.0
5 c 8 3.0
I have a pandas DataFrame that looks similar to the following...
>>> df = pd.DataFrame({
... 'col1':['A','C','B','A','B','C','A'],
... 'col2':[np.nan,1.,np.nan,1.,1.,np.nan,np.nan],
... 'col3':[0,1,9,4,2,3,5],
... })
>>> df
col1 col2 col3
0 A NaN 0
1 C 1.0 1
2 B NaN 9
3 A 1.0 4
4 B 1.0 2
5 C NaN 3
6 A NaN 5
What I would like to do is group the rows of col1 by value and then update any NaN values in col2 to increment in value by 1 based on the last highest value of that group in col1.
So that my expected results would look like the following...
>>> df
col1 col2 col3
0 A 1.0 4
1 A 2.0 0
2 A 3.0 5
3 B 1.0 2
4 B 2.0 9
5 C 1.0 1
6 C 2.0 3
I believe I can use something like groupby on col1 though I'm unsure how to increment the value in col2 based on the last highest value of the group from col1. I've tried the following, but instead of incrementing the value of col1 it updates the value to all 1.0 and adds an additional column...
>>> df1 = df.groupby(['col1'], as_index=False).agg({'col2': 'min'})
>>> df = pd.merge(df1, df, how='left', left_on=['col1'], right_on=['col1'])
>>> df
col1 col2_x col2_y col3
0 A 1.0 NaN 0
1 A 1.0 1.0 1
2 A 1.0 NaN 5
3 B 1.0 NaN 9
4 B 1.0 1.0 4
5 C 1.0 1.0 2
6 C 1.0 NaN 3
Use GroupBy.cumcount only for rows with missing values, add maximum value per group with GroupBy.transform and max and last replace by original values by fillna:
df = pd.DataFrame({
'col1':['A','C','B','A','B','B','B'],
'col2':[np.nan,1.,np.nan,1.,3.,np.nan, 0],
'col3':[0,1,9,4,2,3,4],
})
print (df)
col1 col2 col3
0 A NaN 0
1 C 1.0 1
2 B NaN 9
3 A 1.0 4
4 B 3.0 2
5 B NaN 3
6 B 0.0 4
df = df.sort_values(['col1','col2'], na_position='last')
s = df.groupby('col1')['col2'].transform('max')
df['new'] = (df[df['col2'].isna()]
.groupby('col1')
.cumcount()
.add(1)
.add(s)
.fillna(df['col2']).astype(int))
print (df)
col1 col2 col3 new
3 A 1.0 4 1
0 A NaN 0 2
6 B 0.0 4 0
4 B 3.0 2 3
2 B NaN 9 4
5 B NaN 3 5
1 C 1.0 1 1
Another way:
df['col2_new'] = df.groupby('col1')['col2'].apply(lambda x: x.replace(np.nan, x.value_counts().index[0]+1))
df = df.sort_values('col1')
Question: How do you group a df based on a variable, make a computation using a for loop?
The task is to make a conditional computation based on the value in a column. But the computational constants are dependent upon the value in the reference column. Given this df:
In [55]: df = pd.DataFrame({
...: 'col1' : ['A', 'A', 'B', np.nan, 'D', 'C'],
...: 'col2' : [2, 1, 9, 8, 7, 4],
...: 'col3': [0, 1, 9, 4, 2, 3],
...: })
In [56]: df
Out[56]:
col1 col2 col3
0 A 2 0
1 A 1 1
2 B 9 9
3 NaN 8 4
4 D 7 2
5 C 4 3
I've used the solution here to insert a 'math' column that takes the balance from col3 and adds 10. But now I want to iterate over a list to set the computational variable dependent upon the values in col1. Here's the result:
In [57]: items = ['A', 'D']
In [58]: for item in items:
...: df.loc[:, 'math'] = df.loc[df['col1'] == item, 'col3']
...:
In [59]: df
Out[59]:
col1 col2 col3 math
0 A 2 0 NaN
1 A 1 1 NaN
2 B 9 9 NaN
3 NaN 8 4 NaN
4 D 7 2 2.0
5 C 4 3 NaN
The obvious issue is that the df is over written on each iteration. The math column for index 0 and 1 computed values on the first iteration, but they are removed on the second iteration. The resulting df only considers the last element of the list.
I could go through and add coding to iterate through each index value - but that seems more pathetic than pythonic.
Expected Output for the .mul() example
In [100]: df
Out[100]:
col1 col2 col3 math
0 A 2 0 0.0
1 A 1 1 10.0
2 B 9 9 NaN
3 NaN 8 4 NaN
4 D 7 2 20.0
5 C 4 3 NaN
The problem with your current method is the output of each subsequent iteration overwrites the output of the one before it. So you'd end up with output for just the last item and nothing more.
Select all rows with elements in items and assign, same as you did before.
df['math'] = df.loc[df.col1.isin(items), 'col3'] * 10
Or,
df['math'] = df.query("col1 in #items").col3 * 10
Or even,
df['math'] = df.col3.where(df.col1.isin(items)) * 10
df
col1 col2 col3 math
0 A 2 0 0.0
1 A 1 1 10.0
2 B 9 9 NaN
3 NaN 8 4 NaN
4 D 7 2 20.0
5 C 4 3 NaN
The reason why you fail with assign , cause in each for loop you are assign a Math with new value , like below which will only show the last one and present to the result after the for loop
0 0.0
1 10.0
2 NaN
3 NaN
4 NaN
5 NaN
Name: col3, dtype: float64
0 NaN
1 NaN
2 NaN
3 NaN
4 20.0
5 NaN
Name: col3, dtype: float64
You can do it with below
df.loc[df.col1.isin(items),'math']=df.col3*10
df
Out[85]:
col1 col2 col3 math
0 A 2 0 0.0
1 A 1 1 10.0
2 B 9 9 NaN
3 NaN 8 4 NaN
4 D 7 2 20.0
5 C 4 3 NaN